solve 20.1-1

Example 20.1-2 if needed

20.1-1. Unsteady-State Diffusion in a Thick Slab. Repeat Example 20.1- 2 but use a distribution coefficient K = 0.50 instead of 2.0. Plot the data. Ans. c = c; = 5.75 x 10-2x = 0), c = x 10-2(x = 0.01 m), CLi = 2.87 10-2 kg mol/m3 3 II N a EXAMPLE 20.1-2. Unsteady-State Diffusion in a Semi-infinite Slab A very thick slab has a uniform concentration of solute A of co = 1.0 x 10-2 kg mol A/ m3. Suddenly, the front face of the slab is exposed to a flowing fluid having a concentration c = 0.10 kg mol A/m3 and a convective coefficient ke = 2 x 10-7m/s. The equilibrium 860 distribution coefficient K = Cli/c; = 2.0. Assuming that the slab is a semi-infinite solid, calculate the concentration in the solid at the surface (x = 0) and x = 0.01 m from the surface after t = 3 x 104 s. The diffusivity in the solid is DAB = 4 x 10-m-/s. - Solution: To use Fig. 14.3-3, use Table 20.1-1. Solution: To use Fig. 14.3-3, use Table 20.1-1. 2.0 2 x 10-7 m 202x10515 Kk VDAB -9m' DAB 4 x 10 (3x10^s) = 1.095 4 x 10-9 m2 s For x = 0.01 m from the surface in the solid, . 0.001m 2 2.DAT 0.457 2. || 4 x 10-9m ( 3 x 1045) From the chart, 1 - Y = 0.26. Then, substituting into the equation for (1 - Y) from Table 20.1-1 and solving, - 1x m 1 - Y = C G / - 0.26 C- 1x 10-2 kgmole 10 x 10-2 kgmole - 1x 10-2 kgmole * 2- (= m m C = 2.04 x 10 kgmole/m' (for x = 0.01 m) mo m 1 c = 2.04 x 10-2 kgmole/m? (for x = 0.01 m) For x = 0 m (i.e., at the surface of the solid), X 0 2Dant From the chart, 1 - Y = 0.62. Solving, c = 3.48 x 10-2. This value is the same as ci, as shown in Fig. 20.1-3b. To calculate the concentration Chi in the liquid at the interface, 2.0 3.48 x 10-2 kgmole m 6.96 x 10-2 -2 kgmole Cu = Kc m